An object is placed 12 cm from a firstconverging lens of focal length 13 cm.A se

Question

An object is placed 12 cm from a firstconverging lens of focal length 13 cm.A second converging lens with focal length 39 cm is placed 7.3cm to the right of the first converging lens. (a) Find the position q1 ofthe image formed by the first converging lens.
1 cm

(b) How far from the second lens is the image of the firstlens?
2 cm

(c) What is the value of p2, the objectposition for the second lens?
3cm

(d) Find the position q2 of the image formed bythe second lens.
4

(e) Calculate the magnification of the first lens.
M1 = 5

(f) Calculate the magnification of the second lens.
M2 = 6

(g) What is the total magnification for the system?
Mtotal = 7

(h) Is the final image real or virtual? Isit upright or inverted? 8 real
inverted
upright
noimage
(a) Find the position q1 ofthe image formed by the first converging lens.
1 cm

(b) How far from the second lens is the image of the firstlens?
2 cm

(c) What is the value of p2, the objectposition for the second lens?
3cm

(d) Find the position q2 of the image formed bythe second lens.
4

(e) Calculate the magnification of the first lens.
M1 = 5

(f) Calculate the magnification of the second lens.
M2 = 6

(g) What is the total magnification for the system?
Mtotal = 7

(h) Is the final image real or virtual? Isit upright or inverted? 8 real
inverted
upright
noimage
8 real
inverted
upright
noimage
8 real
inverted
upright
noimage
8

Explanation / Answer

        Given :  Object distance for first lens is : p1 = 12 cm                     Focal length is   f1 =  13 cm We have :                    1 / f1   = 1/ p1 + 1 / q1         or        1/ q1  = 1 / f1 - 1 / p1 = 1 / 13 - 1/ 12     Image formed by first converginglens is :                   q1   = -156.0cm (b)         The second lens isplaced 7.3cm to the right of first lens. Therefore the image formedby the first lens is 19.3cm from the second lens on its right. (c)           Therefore for the second lens p2  =  19.3 cm (d)        We have :               1 / q2 = 1/ f2   - 1/ p2                          = 1/ 39 - 1/ 19.3                  q2    =   -38.20 cm (e)          M1 = - q1 / p1   (e)          M2 = - q2 / p2   (d)           total magnefication is :   M1 * M2   (f)      Final image is real. As thetotal magnification is positive, therefore the image isupright. Hope this helps u! (c)           Therefore for the second lens p2  =  19.3 cm (d)        We have :               1 / q2 = 1/ f2   - 1/ p2                          = 1/ 39 - 1/ 19.3                  q2    =   -38.20 cm (e)          M1 = - q1 / p1   (e)          M2 = - q2 / p2   (d)           total magnefication is :   M1 * M2   (f)      Final image is real. As thetotal magnification is positive, therefore the image isupright. Hope this helps u!

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